Solution for Communication Systems
Solutions Manual for:
Communications Systems, 5th edition by Karl Wiklund, McMaster University, Hamilton, Canada Michael Moher, Space-Time DSP Ottawa, Canada and Simon Haykin, McMaster University, Hamilton, Canada
Published by Wiley, 2009.
Copyright © 2009 John Wiley & Sons, Inc. All Rights Reserved.
Chapter 2 2.1 (a) g (t ) = A cos(2π f c t ) fc = 1 T ⎡ −T T ⎤ t∈⎢ , ⎥ ⎣ 2 2⎦
We can rewrite the half-cosine as: ⎛t ⎞ A cos(2π f c t ) ⋅ rect ⎜ ⎟ ⎝T ⎠ Using the property of multiplication in the time-domain: G ( f ) = G1 ( f ) ∗ G2 ( f )
1 sin(π fT ) [δ ( f − fc ) + δ ( f + fc )] ∗ AT π fT 2 Writing out the convolution: ∞ AT ⎛ sin(πλT ) ⎞ G( f ) = ∫ ⎜ ⎟ [δ (λ − ( f + f c ) + δ (λ − ( f − f c ) ] d λ 2 ⎝ πλT ⎠ −∞ = A ⎛ sin(π ( f + f c )T ) sin(π ( f − f c )T ) ⎞ + ⎜ ⎟ f + fc f − fc 2π ⎝ ⎠ ⎛ ⎞ A ⎜ cos(π fT ) cos(π fT ) ⎟ = − ⎜ 1 ⎟ 2π ⎜ f − 1 ⎟ f+ 2T 2T ⎠ ⎝ = (b)By using the time-shifting property: T exp(− j 2π ft0 ) g (t − t0 ) t0 = 2 ⎛ ⎞ A ⎜ cos(π fT ) cos(π fT ) ⎟ G( f ) = − ⋅ exp(− jπ fT ) ⎜ 1 ⎟ 2π ⎜ f − 1 ⎟ f+ ⎝ 2T 2T ⎠ fc = 1 2T
Copyright © 2009 John Wiley & Sons, Inc. All Rights Reserved.
(c)The half-sine pulse is identical to the half-cosine pulse except for the centre frequency and time-shift.
fc =
1 2Ta
⎡ cos(π fTa ) cos(π fTa ) ⎤ − ⎢ ⎥ ⋅ (cos(π fTa ) − j sin(π fTa )) f + fc ⎦ ⎣ f − fc A ⎡ cos(2π fTa ) cos(2π fTa ) sin(2π fTa ) sin(2π fTa) ⎤ = − +j −j ⎢ ⎥ 4π ⎣ f − f c f + fc f − fc f + fc ⎦ A 2π A ⎡ exp(− j 2π fTa ) exp(− j 2π fTa ) ⎤ − ⎢ ⎥ 4π ⎣ f − fc f + fc ⎦
G( f ) =
=
(d) The spectrum is the same as for (b) except shifted backwards in time and multiplied by -1.
⎛ ⎞ A ⎜ cos(π fT ) cos(π fT ) ⎟ − ⋅ exp( jπ fT ) G( f ) = ⎜ 1 ⎟ 2π ⎜ f − 1 ⎟ f+ 2T 2T ⎠ ⎝ ⎡ ⎤ A ⎢ exp( j 2π fT ) exp( j 2π fT ) ⎥ = − ⎢ ⎥ 1 4π ⎢ f − 1 ⎥ f+ 2T 2T ⎦ ⎣ (e) Because the Fourier transform is a linear operation, this is simply the summation of the results from (b) and (d) ⎡ ⎤ A ⎢ exp( j 2π fT ) + exp(− j 2π fT ) exp( j 2π fT ) + (− j 2π fT ) ⎥ − G(
Communications Systems, 5th edition by Karl Wiklund, McMaster University, Hamilton, Canada Michael Moher, Space-Time DSP Ottawa, Canada and Simon Haykin, McMaster University, Hamilton, Canada
Published by Wiley, 2009.
Copyright © 2009 John Wiley & Sons, Inc. All Rights Reserved.
Chapter 2 2.1 (a) g (t ) = A cos(2π f c t ) fc = 1 T ⎡ −T T ⎤ t∈⎢ , ⎥ ⎣ 2 2⎦
We can rewrite the half-cosine as: ⎛t ⎞ A cos(2π f c t ) ⋅ rect ⎜ ⎟ ⎝T ⎠ Using the property of multiplication in the time-domain: G ( f ) = G1 ( f ) ∗ G2 ( f )
1 sin(π fT ) [δ ( f − fc ) + δ ( f + fc )] ∗ AT π fT 2 Writing out the convolution: ∞ AT ⎛ sin(πλT ) ⎞ G( f ) = ∫ ⎜ ⎟ [δ (λ − ( f + f c ) + δ (λ − ( f − f c ) ] d λ 2 ⎝ πλT ⎠ −∞ = A ⎛ sin(π ( f + f c )T ) sin(π ( f − f c )T ) ⎞ + ⎜ ⎟ f + fc f − fc 2π ⎝ ⎠ ⎛ ⎞ A ⎜ cos(π fT ) cos(π fT ) ⎟ = − ⎜ 1 ⎟ 2π ⎜ f − 1 ⎟ f+ 2T 2T ⎠ ⎝ = (b)By using the time-shifting property: T exp(− j 2π ft0 ) g (t − t0 ) t0 = 2 ⎛ ⎞ A ⎜ cos(π fT ) cos(π fT ) ⎟ G( f ) = − ⋅ exp(− jπ fT ) ⎜ 1 ⎟ 2π ⎜ f − 1 ⎟ f+ ⎝ 2T 2T ⎠ fc = 1 2T
Copyright © 2009 John Wiley & Sons, Inc. All Rights Reserved.
(c)The half-sine pulse is identical to the half-cosine pulse except for the centre frequency and time-shift.
fc =
1 2Ta
⎡ cos(π fTa ) cos(π fTa ) ⎤ − ⎢ ⎥ ⋅ (cos(π fTa ) − j sin(π fTa )) f + fc ⎦ ⎣ f − fc A ⎡ cos(2π fTa ) cos(2π fTa ) sin(2π fTa ) sin(2π fTa) ⎤ = − +j −j ⎢ ⎥ 4π ⎣ f − f c f + fc f − fc f + fc ⎦ A 2π A ⎡ exp(− j 2π fTa ) exp(− j 2π fTa ) ⎤ − ⎢ ⎥ 4π ⎣ f − fc f + fc ⎦
G( f ) =
=
(d) The spectrum is the same as for (b) except shifted backwards in time and multiplied by -1.
⎛ ⎞ A ⎜ cos(π fT ) cos(π fT ) ⎟ − ⋅ exp( jπ fT ) G( f ) = ⎜ 1 ⎟ 2π ⎜ f − 1 ⎟ f+ 2T 2T ⎠ ⎝ ⎡ ⎤ A ⎢ exp( j 2π fT ) exp( j 2π fT ) ⎥ = − ⎢ ⎥ 1 4π ⎢ f − 1 ⎥ f+ 2T 2T ⎦ ⎣ (e) Because the Fourier transform is a linear operation, this is simply the summation of the results from (b) and (d) ⎡ ⎤ A ⎢ exp( j 2π fT ) + exp(− j 2π fT ) exp( j 2π fT ) + (− j 2π fT ) ⎥ − G(